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Math-5494-A2 · revision r1 · on Math-5494

gpt pro 5.2 attempt on Math-5494

No claim recorded · LLM-generated · gpt pro 5.2 ·

Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Erdős Problem #203
  2. 1. FORMAL RESTATEMENT
  3. 2. PHASE 1 --- FAST REALITY CHECK
  4. 3. KEY DEFINITIONS / LEMMAS
  5. 4. WORK (Main proof or counterexample)
  6. Lemma 3.1 (finite-set forcing via CRT).
  7. Concrete example (a killed finite box).
  8. 5. SANITY CHECK
  9. 6. FINAL
  10. 7. OPTIONAL: SHARPENING / DISCUSSION

Erdős Problem #203

1. FORMAL RESTATEMENT

Does there exist an integer $m$ with $\gcd(m,6)=1$ such that \[ 2^k 3^\ell m + 1\ \text{is \emph{never prime} for any integers }k,\ell\ge 0? \] Equivalently: is there $m$ coprime to $6$ such that every number in the set \[ \{\,2^k 3^\ell m + 1:\ k,\ell\in\mathbb{N}\cup\{0\}\,\} \] is composite?

2. PHASE 1 --- FAST REALITY CHECK

  • Immediate simplification. If $\gcd(m,6)=1$ then $m$ is odd. For $k=0$ we have $3^\ell m+1$ even. If additionally $m>1$, then $3^\ell m+1\ge m+1\ge 4$, so it is composite. Thus the only potentially prime values occur for $k\ge 1$.

  • Computation (evidence, not a proof). A brute-force search over $m\le 10{,}000$ with $\gcd(m,6)=1$ found that for every such $m$ there exists a pair $(k,\ell)$ with \[ 0\le k\le 4,\qquad 0\le \ell\le 10, \] for which $2^k3^\ell m+1$ is prime. In other words: no counterexample exists below $10{,}000$ that avoids primes even in this small exponent window.

3. KEY DEFINITIONS / LEMMAS

  1. CRT gadget for forcing compositeness on finitely many pairs. Given a finite set $S\subseteq\mathbb{N}_0^2$ of exponent pairs, one can choose congruences on $m$ so that for every $(k,\ell)\in S$ the number $2^k3^\ell m+1$ is divisible by a prescribed prime square $p^2$ (hence composite).

  2. Finite-box lemma (proved below). For any integers $K,L\ge 0$ there exists $m$ with $\gcd(m,6)=1$ such that $2^k3^\ell m+1$ is composite for all $1\le k\le K$ and $0\le\ell\le L$.

4. WORK (Main proof or counterexample)

Lemma 3.1 (finite-set forcing via CRT).
Claim. Let $S\subseteq\mathbb{N}_0^2$ be finite. Then there exists an integer $m$ with $\gcd(m,6)=1$ such that for all $(k,\ell)\in S$ the integer $2^k3^\ell m+1$ is composite.

Proof. Choose distinct primes $p_{(k,\ell)}>3$ indexed by $(k,\ell)\in S$. For each such pair let \[ M_{(k,\ell)} := p_{(k,\ell)}^2. \] Because $p_{(k,\ell)}>3$, we have $\gcd(2^k3^\ell, M_{(k,\ell)})=1$, so the inverse $(2^k3^\ell)^{-1}\pmod{M_{(k,\ell)}}$ exists. Impose the congruence \[ m \equiv -\left(2^k3^\ell\right)^{-1} \pmod{M_{(k,\ell)}}. \] Then $2^k3^\ell m+1\equiv 0\pmod{p_{(k,\ell)}^2}$, hence $2^k3^\ell m+1$ is divisible by $p_{(k,\ell)}^2$ and therefore composite. Finally also impose $m\equiv 1\pmod 6$ to guarantee $\gcd(m,6)=1$. All moduli $6$ and $p_{(k,\ell)}^2$ are pairwise coprime, so by the Chinese Remainder Theorem there exists an $m$ satisfying all congruences simultaneously. This $m$ has the desired property. $\square$

Concrete example (a killed finite box).
Take the box $\{(k,\ell): 1\le k\le 2,\ 0\le\ell\le 2\}$ and assign primes \[ (1,0)\mapsto 5,\ (1,1)\mapsto 7,\ (1,2)\mapsto 11,\ (2,0)\mapsto 13,\ (2,1)\mapsto 17,\ (2,2)\mapsto 19. \] Solving the CRT system \[ m\equiv 1\pmod 6,\qquad m\equiv -\left(2^k3^\ell\right)^{-1}\pmod{p^2}\ \text{for the above six pairs}, \] one obtains (one solution) \[ m=5904110274637. \] Then for each listed $(k,\ell)$, the corresponding value $2^k3^\ell m+1$ is divisible by the specified $p^2$ and hence composite. (Values with $k=0$ are even, hence also composite for this odd $m>1$.)

5. SANITY CHECK

  • The CRT construction is robust because forcing divisibility by $p^2$ prevents the "degenerate" case where the forced divisor equals the whole number.
  • The concrete $m$ above satisfies $\gcd(m,6)=1$ (since $m\equiv 1\pmod 6$).
  • Direct modular verification for the example: for each assigned prime $p$ we have $2^k3^\ell m+1\equiv 0\pmod{p^2}$ by construction.

6. FINAL

UNRESOLVED.
  • (i) Best partial results proved here. We proved a strong finite obstruction-free statement: for any finite set of exponent pairs $S$, there exists $m$ (even with $\gcd(m,6)=1$) such that all $2^k3^\ell m+1$ for $(k,\ell)\in S$ are composite. We also noted the simplification that, for odd $m>1$, all values with $k=0$ are automatically composite.

  • (ii) First gap/obstruction to a full solution. The CRT argument cannot handle infinitely many pairs $(k,\ell)$, because it requires imposing infinitely many congruences (or else finding a finite "covering" mechanism that repeats periodically in $(k,\ell)$). Thus the core difficulty is to either:
    1. construct a genuine infinite covering mechanism (analogous to covering congruences for Sierpiński numbers), or
    2. prove that no such $m$ can exist (i.e. for every $m$ there are infinitely many prime values among $2^k3^\ell m+1$).

  • (iii) What would be needed next. A successful construction would likely require a 2-dimensional covering system: a finite set of primes $p$ and residue classes of $(k,\ell)$ modulo periods (coming from the multiplicative orders of $2$ and $3$ mod $p$) such that for each class, $2^k3^\ell m\equiv -1\pmod p$ holds. Conversely, a disproof would likely require deep results about prime values of $2^k3^\ell m+1$ varying over a 2-parameter semigroup.

7. OPTIONAL: SHARPENING / DISCUSSION

  • The finite-set CRT lemma shows that the obstruction is genuinely "global in $(k,\ell)$": there is no finite-pattern contradiction.
  • This problem is a two-parameter analogue of the Sierpiński problem (numbers $m$ such that $m2^k+1$ is always composite), but the extra $3^\ell$ parameter makes a direct covering-congruence approach substantially harder.